$8\sin\left(x\right)=3+\frac{4}{\csc\left(x\right)}$
$2^5\cdot2\cdot2^3$
$\left(3x+x^2\right)^3$
$\frac{x^4y^4}{x^4y^4}$
$-4x+7=15$
$y'-y=te^t$
$\frac{1}{x+1}\le1+\frac{2}{x-1}$
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